Without using the formula of equivalent. Find out charge on capacitor and current in all the branches as a function of time.

Explanation

Applying KVL in ABDEA `e-iR=(q)/(2C) rArr i=e/R -q/(2CR)=(2Ce-q)/(2CR)`
`(dq)/(2eC-q)=(dt)/(2CR) rArr underset(0)overset(q)int (dq)/((2eC-q))=(t)/(2CR)`
`(2eC-q)/(2eC)=e^(-t//2RC) rArr q=2eC(1-e^(-t//2RC)`
`q_(1)=q/2=e(1-e^(-t//RC) rArr i_(1)=(e)/(2R)e^(-t//2RC)`
`q_(1)=q/2=eC(1-e^(-t//2RC) rArr i_(2)=(e)/(2R)e^(-t//RC)`
Alternate solution
By equivalent
Time constant of circuit `= 2C xx R =2RC`
maximum charge on capacitor `=2C xx epsi =2Cepsi`
Hence equations of charge and current are as given below
`q=2epsi(1-e^(-t//2RC)`
`q_(1)=q/c=epsiC(1-e^(-t//2RC)`
`rArr i_(1)=epsi_(0)/(2R) e^(-t//2RC)`
`rArr q_(2)=q/2=epsiC(1-e^(-t//2RC)`
`rArr i_(2)=epsi/(2R) e^(-t//RC)`